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1. What the problem is asking (in plain language)

A company has a finite‐population of transactions that are divided into (L) non‑overlapping strata.
For stratum (i) we know

  • the size of the stratum: (N_i)
  • the number of transactions that were inspected: (n_i) (drawn without replacement)
  • the number of inspected transactions that turned out to be “non‑conforming”: (x_i).

The auditor wants a confidence limit (usually an upper bound) for the total number of non‑conforming transactions in the whole population

[ M=\sum_{i=1}^{L} M_i ,\qquad M_i=\hbox{# non‑conforming in stratum }i . ]

The usual textbook solution – treat each stratum as a simple random sample, estimate a variance, and plug a Normal quantile – works only when

  • the within‑stratum sample sizes (n_i) are not tiny,
  • the observed counts (x_i) are not tiny (especially not zero), and
  • the sampling fraction (n_i/N_i) is small (so the finite‑population correction can be ignored).

In many audit situations all three of the above problems occur simultaneously.
The question is: How can we obtain confidence limits that still have (or exceed) the nominal coverage when the normal approximation breaks down?

2. Full step‑by‑step solution

We will build a finite‑population exact (or conservative) interval for the total (M).
The construction proceeds in three logical layers:

  1. Model each stratum as a hypergeometric sampling experiment.
  2. Obtain an exact (Clopper–Pearson) confidence interval for the stratum proportion (p_i=M_i/N_i).
  3. Combine the stratumwise intervals into a confidence interval for the total (M=\sum N_i p_i).

The final interval is guaranteed to have at least the nominal coverage (e.g., 95 %).
If a slightly less conservative interval is desired, a mid‑(P) or exact unconditional approach can be used (described briefly in the “Refinements” paragraph).

2.1 Hypergeometric model for a single stratum

In stratum (i) the unknown number of non‑conforming units is (M_i) (an integer between 0 and (N_i)).
A simple random sample of size (n_i) without replacement yields

[ X_i\mid M_i\;\sim\;\operatorname{Hypergeometric}\bigl(N_i,\;M_i,\;n_i\bigr), \qquad \Pr{X_i=x\mid M_i}= \frac{\binom{M_i}{x}\binom{N_i-M_i}{\,n_i-x\,}}{\binom{N_i}{n_i}} . ]

The observed value is (x_i). The goal is to infer the unknown integer (M_i).

2.2 Exact (Clopper–Pearson) confidence interval for (M_i)

The Clopper–Pearson (also called “exact”) interval for a binomial proportion is obtained by inverting a one‑sided hypothesis test.
The same idea works for the hypergeometric distribution, except that the parameter is the integer (M_i).

For a nominal one‑sided level (\alpha) (e.g., (\alpha=0.05) for a 95 % upper bound) we define

  • Upper confidence limit (U_i) as the smallest integer (m) such that

[ \Pr{X_i\le x_i\mid M_i=m}\;\ge\;1-\alpha . \tag{1} ]

  • Lower confidence limit (L_i) as the largest integer (m) such that

[ \Pr{X_i\ge x_i\mid M_i=m}\;\ge\;1-\alpha . \tag{2} ]

Because the hypergeometric cdf can be evaluated exactly (most statistical packages have phyper/dhyper), the limits are found by a simple binary search over the integer range ([0,N_i]).

When (x_i=0) the lower limit is trivially (L_i=0); the upper limit is the smallest (m) satisfying (1).
When (x_i=n_i) the upper limit is (U_i=N_i); the lower limit is the largest (m) satisfying (2).

These limits are conservative: the true coverage is at least (1-\alpha).

2.3 From stratum proportions to a total‑population interval

Define the stratum‑wise proportion interval

[ \frac{L_i}{N_i}\; \le\; p_i\; \le\; \frac{U_i}{N_i}. ]

Because the strata are independent samples (sampling without replacement in each stratum does not affect the others), a simultaneous confidence statement can be obtained by the Bonferroni correction:

  • Choose a per‑stratum error probability (\alpha_i = \alpha/L).
  • Compute the ((1-\alpha_i)) Clopper–Pearson interval ([L_i,U_i]) for each stratum.

Then

[ \Pr\Bigl{ L_i\le M_i \le U_i\;\text{ for all } i=1,\dots ,L\Bigr}\;\ge\;1-\alpha . \tag{3} ]

Finally, add the lower (resp. upper) bounds across strata:

[ \boxed{ \; M_{\text{L}} \;=\; \sum_{i=1}^{L} L_i, \qquad M_{\text{U}} \;=\; \sum_{i=1}^{L} U_i \;} ]

and the interval

[ \bigl[M_{\text{L}},\;M_{\text{U}}\bigr] ]

is a (100(1-\alpha)\%) confidence interval for the total number of non‑conforming transactions.

Because each (L_i) and (U_i) is an integer, the total bounds are also integers, which is appropriate for a count.

2.4 Worked numerical example

Assume three strata:

Stratum (i) (N_i) (n_i) (x_i)
1 500 30 0
2 800 40 2
3 200 25 1

Nominal 95 % overall confidence → (\alpha = 0.05).
Bonferroni per‑stratum error: (\alpha_i = 0.05/3 \approx 0.0166667).

Stratum 1 ( (x_1=0) )

We need the smallest (m) such that

[ \Pr{X_1\le 0\mid M_1=m}= \frac{\binom{N_1-m}{n_1}}{\binom{N_1}{n_1}} \ge 1-\alpha_i = 0.98333 . ]

A quick search gives (m=5).
Thus ([L_1,U_1]=[0,5]).

Stratum 2 ( (x_2=2) )

Using a computer routine (e.g., phyper in R) we find

  • Lower limit (L_2=1) (largest (m) with (\Pr{X\ge2\mid M=m}\ge0.98333)).
  • Upper limit (U_2=13) (smallest (m) with (\Pr{X\le2\mid M=m}\ge0.98333)).

So ([L_2,U_2]=[1,13]).

Stratum 3 ( (x_3=1) )

Similarly,

  • Lower limit (L_3=0).
  • Upper limit (U_3=6).

Thus ([L_3,U_3]=[0,6]).

Combine

[ M_{\text{L}} = L_1+L_2+L_3 = 0+1+0 = 1, \qquad M_{\text{U}} = U_1+U_2+U_3 = 5+13+6 = 24 . ]

Hence a conservative 95 % confidence interval for the total number of non‑conforming transactions is

[ \boxed{ \;1 \;\le\; M \;\le\; 24\; } . ]

If the auditor only needs an upper bound, the interval reduces to (M\le 24).

2.5 Refinements (less conservative alternatives)

Refinement Idea When useful
Mid‑(P) Clopper–Pearson Replace the tail probability in (1)–(2) by (\frac12) of the probability of the observed count (the “mid‑(P)” adjustment). Gives intervals that are closer to the nominal level while keeping good coverage.
Exact unconditional interval (e.g., “exact test inversion” for the sum) Compute the exact distribution of (X=\sum_i X_i) under every possible total (M) (using convolution of hypergeometrics) and find the set of (M) that are not rejected by a two‑sided test at level (\alpha). Gives the shortest possible exact interval; computationally intensive but feasible for modest (L) and (N_i).
Wilson or Agresti–Coull with finite‑population correction Approximate each stratum proportion by a Wilson interval, then apply the FPC factor (\sqrt{(N_i-n_i)/(N_i-1)}). Works well when (n_i) is not extremely small (say (n_i\ge5)) and the sampling fraction is moderate.
Bootstrap (finite‑population) Resample without replacement from each stratum using the observed data, recompute (\hat M) many times, and take empirical quantiles. Gives a data‑driven interval; coverage must be checked by simulation for the specific design.

For most audit applications the Bonferroni‑combined Clopper–Pearson interval is both simple to implement and guaranteed to be at least as wide as needed, which satisfies the “defensible” requirement.

3. Final answer (concise statement)

Model each stratum as a hypergeometric sample. For each stratum compute an exact one‑sided Clopper–Pearson confidence limit for the unknown number of non‑conforming items (M_i) at per‑stratum error (\alpha/L). Add the lower (or upper) limits across strata. The resulting integer interval ([\,\sum L_i,\;\sum U_i\,]) is a valid ((1-\alpha)) confidence interval for the total number of non‑conforming transactions (M). The method works for tiny samples, zero counts, and large sampling fractions because it does not rely on a Normal approximation.

4. Common Mistakes

Mistake Why it is wrong How to avoid it
Using the normal approximation (\hat M_i\pm z_{1-\alpha/2}\sqrt{\hat V_i}) when many (x_i) are 0 or 1. The normal distribution is a poor fit to a highly skewed (or point‑mass) hypergeometric count, leading to under‑coverage. Switch to exact (Clopper–Pearson) intervals whenever any stratum has (x_i\le5) or (n_i/N_i>0.1).
**Pooling all strata together and applying a    

Original question: How to construct confidence limits based on small stratified samples of finite populations? on Cross Validated (Stats Stack Exchange), licensed CC BY-SA.